Summation to n of kth Harmonic Number over k+1

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Theorem

$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1} = \dfrac { {H_{n + 1} }^2 - \harm 2 {n + 1} } 2$

where:

$H_n$ denotes the $n$th harmonic number
$\harm 2 n$ denotes the general harmonic number of order $2$ evaluated at $n$.


Proof

\(\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \dfrac 1 {k + 1} \paren {H_{k + 1} - \dfrac 1 {\paren {k + 1} } }\) Definition of Harmonic Number
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {\dfrac {H_{k + 1} } {k + 1} - \dfrac 1 {\paren {k + 1} \paren {k + 1} } }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \dfrac {H_{k + 1} } {k + 1} - \sum_{k \mathop = 1}^n \dfrac 1 {\paren {k + 1}^2}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 2}^{n + 1} \dfrac {H_k} k - \sum_{k \mathop = 2}^{n + 1} \dfrac 1 {k^2}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \dfrac {H_1} 1 - \paren{\sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2} - \dfrac 1 {1^2} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - 1 - \paren{\sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2} - 1}\) Harmonic Number $H_1$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2}\) simplifying
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \harm 2 {n + 1}\) Definition of General Harmonic Numbers
\(\ds \) \(=\) \(\ds \dfrac { {H_{n + 1} }^2 + \harm 2 {n + 1} } 2 - \harm 2 {n + 1}\) Summation to n of kth Harmonic Number over k
\(\ds \) \(=\) \(\ds \dfrac { {H_{n + 1} }^2 + \harm 2 {n + 1} - 2 \harm 2 {n + 1} } 2\)
\(\ds \) \(=\) \(\ds \dfrac { {H_{n + 1} }^2 - \harm 2 {n + 1} } 2\)


Hence the result.

$\blacksquare$


Sources

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